NCERT Plus One (Class 11) Mathematics • Chapter 8

Sequences and Series

Q: Find the 5th term of the sequence defined by aₙ = (−1)ⁿ⁻¹ · 5ⁿ⁺¹.

Substitute n = 5 : a₅ = (−1)⁵⁻¹ · 5⁵⁺¹ = (−1)⁴ · 5⁶ = 15,625 .

Q: Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.

a₈ = a · r⁷ = a · 2⁷ = 192 ⇒ a = 192 128 = 3 2 . Therefore a₁₂ = a · r¹¹ = 3 2 · 2¹¹ = 3 · 2¹⁰ = 3,072 .

Q: How many terms of the G.P. 3, 3², 3³, ... are needed to give the sum 120?

Here a = 3, r = 3 . Sₙ = 3(3ⁿ − 1) 3 − 1 = 120 ⇒ 3(3ⁿ − 1) = 240 ⇒ 3ⁿ − 1 = 80 ⇒ 3ⁿ = 81 = 3⁴ ⇒ n = 4 .

Q: The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.

Take terms as a r , a, ar . Product: a³ = 1 ⇒ a = 1 . Sum: 1 r + 1 + r = 39 10 ⇒ 10r² − 29r + 10 = 0 ⇒ (5r − 2)(2r − 5) = 0 . r = 2 5 or 5 2 → Terms: 5/2, 1, 2/5 (or 2/5, 1, 5/2 ).

Q: Find the sum to n terms of the sequence 8, 88, 888, 8888, ...

Sₙ = 8(1 + 11 + 111 + ...) = 8 9 (9 + 99 + 999 + ...) = 8 9 [(10 − 1) + (10² − 1) + ... + (10ⁿ − 1)] = 8 9 [ 10(10ⁿ − 1) 9 − n] = 80 81 (10ⁿ − 1) − 8 9 n .

Q: Insert 3 numbers between 1 and 256 so that the resulting sequence is a G.P.

Let 1, G₁, G₂, G₃, 256 be in G.P. (5 terms total). 256 = 1 · r⁴ ⇒ r = ±4 . • For r = 4 : 4, 16, 64 . • For r = −4 : −4, 16, −64 .

Q: If A.M. and G.M. of roots of a quadratic equation are 8 and 5 respectively, obtain the quadratic equation.

Let roots be α, β . A.M. = α + β 2 = 8 ⇒ α + β = 16 . G.M. = √(αβ) = 5 ⇒ αβ = 25 . Quadratic equation: x² − (α + β)x + αβ = 0 ⇒ x² − 16x + 25 = 0 .

Pattern sequences, series, Geometric Progression (nth term & sum) & AM ≥ GM inequality.

📄 Official NCERT Textbook PDF

Topic 8.2: Sequences as Pattern Functions & Fibonacci Recurrence

Count the petals on a lily (3), buttercup (5), delphinium (8), marigold (13), or daisy (21) — nature runs on 1, 1, 2, 3, 5, 8, 13, 21, ...! Each number equals the sum of the two before it! An ordered list following a rule — where position 1, 2, 3, ... determines the value — is a Sequence!

WHAT IS A SEQUENCE?Ordered list a₁, a₂, a₃, ... — formally a function with domain ℕ!

EXPLICIT RULEPlug n directly! E.g. aₙ = 2n + 5 ⇒ 7, 9, 11, ...

FIBONACCI RULEa₁ = 1, a₂ = 1, and aₙ = aₙ₋₁ + aₙ₋₂ (for n > 2)

🧠 Memory Trick: A Sequence is an ordered line with COMMAS (a₁, a₂, a₃), where position n MUST be a positive whole number (1, 2, 3, ...)!

💡 Simple Kid-Check: Think of a sequence like numbered train coaches: Coach #1 holds a₁, Coach #2 holds a₂, Coach #n holds aₙ. You NEVER have Coach #0 or Coach #2.5!

1
Finite vs Infinite Sequence: Stops at a last term (2, 4, 6, 8) vs goes on forever (1, 1, 2, 3, 5, ...).
2
Alternating Signs: Powers like (−1)ⁿ or (−1)ⁿ⁻¹ flip signs back and forth between + and −!

🧒 Quick Term Check: Find the first 4 terms of aₙ = (−1)ⁿ⁻¹ · 5ⁿ⁺¹.
• n = 1: (−1)⁰ · 5² = +25
• n = 2: (−1)¹ · 5³ = −125
• n = 3: (−1)² · 5⁴ = +625
• n = 4: (−1)³ · 5⁵ = −3,125!

Watch TWO classic 1-mark traps: (1) Writing (−1)ⁿ vs (−1)ⁿ⁻¹: at n = 1, (−1)¹ = −1 (starts negative!), whereas (−1)¹⁻¹ = (−1)⁰ = +1 (starts positive!). (2) In recurrence relations like Fibonacci, you CANNOT compute aₙ directly without first computing earlier terms!

Topic 8.3: Series & Sigma (Σ) Summation Idea

Suppose you save pocket money in a jar: ₹10 on Monday, ₹20 on Tuesday, ₹40 on Wednesday. The list 10, 20, 40 is a Sequence. Now ask "How much TOTAL money sits in the jar?" — change commas into + signs: 10 + 20 + 40 = ₹70! That added expression is a Series!

SERIES & SIGMA (Σ)Sum expression a₁ + a₂ + ... + aₙ written compactly as ∑_k=1ⁿ aₖ

PARTIAL SUM SₙSₙ = a₁ + a₂ + ... + aₙ (total of first n terms!)

RECOVER TERM aₙaₙ = Sₙ − Sₙ₋₁ (for n ≥ 2) and a₁ = S₁!

🧠 Memory Trick: Sequence = Commas ( , ); Series = Pluses ( + )! And the Greek letter Σ (Sigma) literally means "Sum everything from bottom index to top index"!

💡 Simple Kid-Check: How do you find JUST the 5th day's deposit a₅ if you only know the running piggy-bank total Sₙ? Take the total after 5 days (S₅) minus the total after 4 days (S₄): a₅ = S₅ − S₄!

Cool Odd-Number Series Discovery:
1 = 1 = 1² | 1 + 3 = 4 = 2² | 1 + 3 + 5 = 9 = 3² | ∑_k=1ⁿ (2k − 1) = n²!

🧒 Quick Term Recovery Example: If the sum of first n terms is Sₙ = n² + 2n, find the 4th term a₄.
• S₄ = 4² + 2(4) = 16 + 8 = 24.
• S₃ = 3² + 2(3) = 9 + 6 = 15.
• Therefore a₄ = S₄ − S₃ = 24 − 15 = 9!

NEVER plug n = 1 into aₙ = Sₙ − Sₙ₋₁ blindly if S₀ ≠ 0 (or when the formula isn't defined at 0)! Always compute the very first term directly using a₁ = S₁, and use Sₙ − Sₙ₋₁ only for n ≥ 2! Also remember: a Series is the formal sum 1 + 3 + 5, NOT only its final number 9!

Topic 8.4.1: Geometric Progression (G.P.) & Common Ratio r

Drop a super-bouncy rubber ball from 81 cm and each bounce rebounds to 2/3 of its previous height: 81, 54, 36, 24, .... Or watch a WhatsApp message double to 2 friends, then 4, 8, 16! Whenever every step MULTIPLIES by the exact same non-zero constant r, you have a Geometric Progression (G.P.)!

G.P. TESTRatio aₙ₊₁aₙ = r is constant everywhere (a ≠ 0, r ≠ 0!)

STANDARD FORMa, ar, ar², ar³, ... (first term a, common ratio r)

3 TERMS IN G.P.When product is given, pick ar, a, ar (r cancels on multiply!)

🧠 Memory Trick: Common Ratio r = Second TermFirst Term (ALWAYS Next ÷ Previous, NEVER Previous ÷ Next!)

💡 Simple Kid-Check: Watch how r shapes the G.P.: if r > 1 (like r = 2), terms explode upward! If 0 < r < 1 (like r = 1/2), terms shrink toward zero! If r < 0 (like r = −3), signs flip + − + − every step!

🔎 Quick G.P. Check: Is 16, −8, 4, −2, ... a G.P.?
Check ratios: −816 = −12, 4−8 = −12, −24 = −12.
Every ratio matches! So YES, it's a G.P. with a = 16 and r = −1/2!

🎯 Smart 3-Mark Trick (a/r, a, ar): Product of 3 G.P. terms is 216.
Multiply (ar)(a)(ar) = a³ = 216 ⇒ a = 6 instantly without solving for r yet!

Watch THREE G.P. traps: (1) Flipping ratio upside-down (a₁ ÷ a₂ instead of a₂ ÷ a₁): in 16, 8, 4, r = 8/16 = 1/2 (NOT 2!). (2) In "−2/7, x, −7/2 are in G.P.", solving x² = 1 gives BOTH +1 AND −1 — forgetting −1 loses half a mark! (3) Neither a nor r can ever be 0 in a standard G.P.!

Topic 8.4.2: General Term (a·rⁿ⁻¹) & Finite Sum Sₙ of a G.P.

Remember the clever sage who asked the king for 1 grain of rice on square 1 of a chessboard, 2 on square 2, 4 on square 3, up to square 64? Square 64 alone holds 2⁶³ grains — and the TOTAL across all squares equals 2⁶⁴ − 1! Two tidy G.P. formulas handle ANY term and ANY finite sum in seconds!

nᵗʰ TERM OF G.P.aₙ = a · rⁿ⁻¹ (Exponent is n − 1, NOT n!)

FINITE SUM (r ≠ 1)Sₙ = arⁿ − 1r − 1 (r > 1) | a1 − rⁿ1 − r (r < 1)

SPECIAL CASE r = 1If r = 1: a, a, ..., a → Sₙ = n · a!

🧠 Memory Trick: For the classic board-exam sum 8 + 88 + 888 + ..., pull out 8/9 to convert every term into (10 − 1) + (10² − 1) + ..., then apply G.P. sum!

💡 Simple Kid-Check: Why does aₙ = a · rⁿ⁻¹ have (n − 1)? Because you start at a₁ = a and only take n − 1 jumps to reach step n!

🏆 Classic 3-Mark DHSE Sum: Find sum to n terms of 8 + 88 + 888 + 8888 + ...:
1) Factor 8 out, multiply & divide by 9: Sₙ = 89(9 + 99 + 999 + ...)
2) Rewrite as 10-powers: = 89[(10 − 1) + (10² − 1) + ... + (10ⁿ − 1)]
3) Group G.P. + ones: = 89[10(10ⁿ − 1)9 − n] = 8081(10ⁿ − 1) − 89n!

Watch THREE common exam errors: (1) Writing aₙ = a·rⁿ instead of a·rⁿ⁻¹! (2) Plugging r = 1 into (1 − rⁿ)/(1 − r) causing 0/0 — ALWAYS check r = 1 separately (Sₙ = n·a)! (3) In 8 + 88 + 888 + ... forgetting the −n inside the bracket (those n copies of −1 add to −n, NOT −1!).

Topic 8.4.3: Geometric Mean (G = √ab) & Inserting k G.M.s

Suppose a plant grows from 4 cm to 36 cm in two equal multiplicative jumps: 4 → G → 36! What height was it halfway through? Not (4+36)/2 = 20 (that's additive!) — we need the SAME multiplication factor r each step: 4 × 3 = 12, and 12 × 3 = 36! So G = √(4 × 36) = 12 is the Geometric Mean!

GEOMETRIC MEAN (G)Between positive a, b: G = √(ab) (so a, G, b form a G.P.!)

INSERT k G.M.sa, G₁, G₂, ..., Gₖ, b has k + 2 terms total → b = a · rᵏ⁺¹!

RATIO FORMULAr = (ba)^1k+1, then G₁ = ar, G₂ = ar², ..., Gₖ = arᵏ

🧠 Memory Trick: Inserting k numbers between a and b makes b sit at position k + 2 — so its exponent on r is k + 1 (NEVER k)!

💡 Simple Kid-Check: Geometrically, a rectangle of width a and height b has area ab. The side length of a square having the EXACT same area ab is G = √(ab) — that's literally why it's called Geometric Mean!

🧒 Classic 3-Mark Inserter: Insert 3 numbers between 1 and 256 so the resulting sequence is a G.P.:
• List: 1, G₁, G₂, G₃, 256 → 5 terms total (k + 2 = 5).
• 5th term: a₅ = 1 · r⁴ = 256 ⇒ r = ±4.
• For r = 4: inserted numbers are 4, 16, 64.
• For r = −4: inserted numbers are −4, 16, −64!

Watch TWO classic exam traps: (1) Using b = a·rᵏ instead of b = a·rᵏ⁺¹ when inserting k G.M.s! Always count total terms (1 + k + 1 = k + 2) so last term b is a·rᵏ⁺¹. (2) When k is odd (so k + 1 is EVEN, like r⁴ = 256), remember EVEN roots yield BOTH +r and −r unless the question restricts to positive numbers!

Topic 8.5: Relationship Between A.M. & G.M. (A.M. ≥ G.M. & Proofs)

Compare the two averages of 4 and 16: Arithmetic Mean A = (4 + 16)/2 = 10, while Geometric Mean G = √(4 × 16) = 8. Notice 10 ≥ 8! Try ANY two positive numbers you like — A.M. is ALWAYS greater than or equal to G.M., and they tie ONLY when both numbers are equal!

GOLDEN INEQUALITYFor positive a, b: A ≥ G, i.e., a + b2 ≥ √(ab)

1-LINE PROOFA − G = (√a − √b)²2 ≥ 0 (Square ≥ 0; equal ⇔ a = b!)

QUADRATIC FINDERGiven A, G, numbers a, b are roots of x² − 2Ax + G² = 0

🧠 Memory Trick: Check minimum value traps in 1 second! For positive x, what is the minimum of x + 1/x? Set A.M. ≥ G.M.: (x + 1/x)/2 ≥ √(x · 1/x) = 1 ⇒ x + 1/x ≥ 2!

💡 Simple Kid-Check: Look at a semicircle with diameter a + b. The circle's radius from center to top is A = (a + b)/2. Any vertical pole inside the semicircle has length G = √(ab). Obviously a vertical pole can NEVER poke above the circle's radius — so A ≥ G visually!

🏆 Classic 4-Mark DHSE Problem: If A.M. and G.M. of two positive numbers are 10 and 8, find the numbers.
1) A = a + b2 = 10 ⇒ a + b = 20.
2) G = √(ab) = 8 ⇒ ab = 64.
3) Build quadratic x² − (a + b)x + ab = 0 ⇒ x² − 20x + 64 = 0 ⇒ (x − 4)(x − 16) = 0.
So the two numbers are 4 and 16!

Watch TWO sneaky traps: (1) Building the quadratic as x² − Ax + G = 0 instead of x² − 2Ax + G² = 0! Remember: sum of roots = a + b = 2A and product = ab = G²! (2) A.M. ≥ G.M. requires BOTH numbers a, b to be positive (a, b > 0) — never apply it blindly to negative numbers!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 1 Marks): Find the 5th term of the sequence defined by aₙ = (−1)ⁿ⁻¹ · 5ⁿ⁺¹.

Substitute n = 5:
a₅ = (−1)⁵⁻¹ · 5⁵⁺¹ = (−1)⁴ · 5⁶ = 15,625.

Q2 (Kerala HSE Say • 1 Marks): Find the Geometric Mean of 4 and 9.

G = √(4 × 9) = √36 = 6.

Q3 (Kerala HSE March • 2 Marks): Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.

a₈ = a · r⁷ = a · 2⁷ = 192 ⇒ a = 192128 = 32.
Therefore a₁₂ = a · r¹¹ = 32 · 2¹¹ = 3 · 2¹⁰ = 3,072.

Q4 (Kerala HSE Say • 2 Marks): How many terms of the G.P. 3, 3², 3³, ... are needed to give the sum 120?

Here a = 3, r = 3.
Sₙ = 3(3ⁿ − 1)3 − 1 = 120 ⇒ 3(3ⁿ − 1) = 240 ⇒ 3ⁿ − 1 = 80 ⇒ 3ⁿ = 81 = 3⁴ ⇒ n = 4.

Q5 (Kerala HSE March • 3 Marks): The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.

Take terms as ar, a, ar.
Product: a³ = 1 ⇒ a = 1.
Sum: 1r + 1 + r = 3910 ⇒ 10r² − 29r + 10 = 0 ⇒ (5r − 2)(2r − 5) = 0.
r = 25 or 52 → Terms: 5/2, 1, 2/5 (or 2/5, 1, 5/2).

Q6 (Kerala HSE March • 3 Marks): Find the sum to n terms of the sequence 8, 88, 888, 8888, ...

Sₙ = 8(1 + 11 + 111 + ...) = 89(9 + 99 + 999 + ...)
= 89[(10 − 1) + (10² − 1) + ... + (10ⁿ − 1)]
= 89[10(10ⁿ − 1)9 − n] = 8081(10ⁿ − 1) − 89n.

Q7 (NCERT / DHSE Model • 3 Marks): Insert 3 numbers between 1 and 256 so that the resulting sequence is a G.P.

Let 1, G₁, G₂, G₃, 256 be in G.P. (5 terms total).
256 = 1 · r⁴ ⇒ r = ±4.
• For r = 4: 4, 16, 64.
• For r = −4: −4, 16, −64.

Q8 (Kerala HSE Say • 4 Marks): If A.M. and G.M. of roots of a quadratic equation are 8 and 5 respectively, obtain the quadratic equation.

Let roots be α, β.
A.M. = α + β2 = 8 ⇒ α + β = 16.
G.M. = √(αβ) = 5 ⇒ αβ = 25.
Quadratic equation: x² − (α + β)x + αβ = 0 ⇒ x² − 16x + 25 = 0.